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Trying to prove by contrary that if $f$ is an onto linear transformation, $g$ is any function, and $h=g \circ f$ is linear, then $g$ must be a linear function.

    f:Rn->Rm
    g:Rm->Rp
    h:Rn->Rp

3 Answers 3

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Sorry I misread question. What is described in the previous edits is not an approach to the actual question asked.

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    In this equation, x would be a vector, a would be a scalar and b would just be some random constant? Correct me if I am wrong, the problem gives us the fact that **h(~x)=g(f(~x))**, where **~x** is a vector, and I am not sure if that changes your equation at all – 2017-02-12
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    Correct , $b$ is a vector as well. I used what the problem gave us in writing down $h(ax+b) = g(f(ax+b))$. What is left is to use the properties that $f$ and $g$ are linear to show that $g(f(ax+b))$ is $ah(x) + h(b)$. We then will have shown that $h$ is linear. – 2017-02-12
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    @stableMatch: read the question! $g$ is not linear. – 2017-02-12
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    Ah, you are correct @RobArthan. Chill out though – 2017-02-12
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    Well @RobArthan I was trying to prove that _g_ must be linear by assuming the contrary of _g_ being non-linear, but stableMatch has given me a better way to prove it, imo. – 2017-02-12
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If $f$ is linear and onto and $f \circ g$ is linear then $g$ must be linear. If $f$ is linear and onto, you can pull a counter-example to the linearity of $g$ back to a counter-example of the linearity of $g \circ f$. E.g., if $g(a + b) \neq g(a) + g(b)$, choose $x$ and $y$ such that $f(x) = a$ and $f(y) = b$ and then you will have

$$(g \circ f)(x + y) = g(a + b) \neq g(a) + g(b) = (g \circ f)(x) + (g \circ f)(y). $$

And similarly if $g(\lambda a) \neq \lambda g(a)$ for some $\lambda$ and $a$.

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    Ok, and so b/c _g(f(x+y)) ≠ g(f(x))+g(f(y))_, then _h(x)_ which equals _g(f(x))_ must be non-linear. But b/c it is stated that _h(x)_ is linear, then _g_ must be linear. – 2017-02-12
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    Yes, but "b/c" is not a standard abbreviation for "because" in MSE (so I have been staring at your comment for a minute or so wondering what it meant). – 2017-02-12
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    Ohh sorry, used to just text style. Will make sure not to use that on a formal proof! – 2017-02-12
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    No worries - the problem is that $b/c$ looks like a fraction when we're writing mathematics! – 2017-02-12
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    I don't mean to pester, but I have another proof problem that I have asked on here. It's about spans and linear independence. The question is in my profile, and I believe I am on the right track, I am just stuck on the equation set-up. If you are knowledgeable in linear algebra, taking a look would be greatly appreciated. – 2017-02-13
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    If you have another question, post it as a question. – 2017-02-13
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Be $f:U\to V$ onto and linear, and $g:V\to W$ so that $h=g\circ f$ is linear, where $U$, $V$, $W$ are vector spaces over the field $K$.

Now be $v,w\in V$ and $\alpha,\beta\in K$. Then because $f$ is onto, there are $t,u\in U$ such that $f(t)=v$ and $f(u)=w$. Since $f$ is linear, this implies that $f(\alpha t+\beta u) = \alpha v+\beta w$. Thus we have \begin{align} g(\alpha v + \beta w) &= g(f(\alpha t+\beta u))\\ &= h(\alpha t+\beta u) && \text{definition of $h$}\\ &= \alpha h(t) + \beta h(u) && \text{linearity of $h$}\\ &= \alpha g(f(t)) + \beta g(f(u)) && \text{definition of $h$}\\ &= \alpha g(v) + \beta g(w) \end{align} Thus $g$ is a linear function.